Problem: $\sum\limits_{n=0}^{\infty }{{\left( -1 \right)}^{n+1}}\frac{{{x}^{2n+3}}}{\left( 2n \right)!}~~$ is the Maclaurin series for which function? Choose 1 answer: Choose 1 answer: (Choice A) A $x^3\cos (-x)$ (Choice B) B $-x^3\cos x$ (Choice C) C $\cos x^3$ (Choice D) D $-\cos x^3$ (Choice E) E $\sin x^3$ (Choice F) F $-x^3\sin(-x)$
Note that the terms of the given series alternate in sign and have only even factorials in the denominator. This suggests that we consider the Maclaurin series for the function $~\cos x\,$. $ \cos~ x=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n}}}{\left( 2n \right)!}+...$ We need the sign of each term to be changed, and we need each of the powers of $~x~$ to be larger by 3. Hence we have $ -x^3 \cos~ x=-x^3+\frac{{{x}^{5}}}{2!}-\frac{{{x}^{7}}}{4!}-...+{{\left( -1 \right)}^{n+1}}\frac{{{x}^{2n+3}}}{\left( 2n \right)!}+...\,$